Algebra help please!?

Question

I kind of have the answers for these two questions, but I'm just not sure. Please help!

A. (Solve using the quadratic formula)

2x^2+2x-5=0


B. (## is square root, so #4# is square root of 4)

#3x+1#+1=x

Any help would be appreciated. Thanks in advance!

Ok, I know the quadritic formula, I just need to make sure my answer is correct. The second question: sq rt of (3x+1) +1=x hope that clears things up :)

Answer 1

Using quadratic formula, x = [-2 +/- #(-2_)^2- 4(2)(-5)#]/2(2)

=( -2 +/- #44#)/4 =(-2 +/- 2#11#)/4

It can be simplified to (-1 +/- #11#)/2

For the second one, first move the1 to the other side:

#3x+1# = x -1. Then square both sides:

3x + 1 = x^2 -2x +1. Get a zero on one side, then solve with quadratic formula or factoring (if possible).

0 = x^2 - 5x
0 = x(x-5)
x = {0, 5)

Answer 2

quadratic formula for quadratic eq: ax^2+bx+c =0
is (-b (+or-) #(b^2 -4ac)#)/2a

so solve first

for second
#(3x+1)# +1 =x
#3x+1# = x-1
3x+1 = (x-1)^2
3x +1 = x^2 -2x +1
x^2 -5x =0
x(x-5)=0
hence the roots are: x=0 and x =5

Answer 3

Find determinant (b^2-4*a*c)

Then the solutions would be (-b+sqrt(determinant))/2 and
(-b-sqrt(determinant))/2

Second qn is not clear

Answer 4

-b +/- sqrt(b^2-4ac)
-------------------------
2a

A. (-1 +sqrt(11))/2 or (-1-sqrt(11))/2

B. x^2 - 5x; x=0 or x=5